Home / AP® Exam / AP® Physics C: Electricity and Magnetism / AP Physics C E&M – 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law- Study Notes

AP Physics C E&M- 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law- Study Notes- New Syllabus

AP Physics C E&M- 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law – Study Notes

AP Physics C E&M- 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law – Study Notes – per latest Syllabus.

Key Concepts:

  • Magnetic Field Produced by a Current-Carrying Wire
  • Force on Current-Carrying Wires in a Magnetic Field

AP Physics C E&M-Concise Summary Notes- All Topics

Magnetic Field Produced by a Current-Carrying Wire

A moving charge or current produces a magnetic field in the surrounding space. The direction of the magnetic field around a straight current-carrying wire is determined by the right-hand rule: if the thumb points in the direction of current, the curled fingers show the circular magnetic field lines.

Biot–Savart Law (General Expression):

The magnetic field at a point due to a small current element is given by:

\( \delta \vec{B} = \dfrac{\mu_0}{4\pi} \dfrac{I \, \delta \vec{l} \times \hat{r}}{r^2} \)

  • \(\mu_0\): permeability of free space \((4\pi \times 10^{-7} \, \mathrm{T \cdot m/A})\).
  • \(I\): current in the wire.
  • \(\delta \vec{l}\): small element of the wire in the direction of current.
  • \(\hat{r}\): unit vector pointing from the current element to the field point.
  • \(r\): distance between the current element and the field point.

Magnetic Field Produced by a Current-Carrying Wire

An electric current consists of moving charges, and these moving charges produce a magnetic field around the wire. The strength and direction of this magnetic field depend on the current and the geometry of the wire.

Right-Hand Rule:

  • Point the thumb of your right hand in the direction of the conventional current (\(I\)).
  • The curl of your fingers shows the circular direction of the magnetic field lines around the wire.

Magnetic Field Around a Long Straight Wire:

Using Ampère’s law or the Biot–Savart law, the magnetic field at a perpendicular distance \(r\) from a long straight wire carrying current \(I\) is:

\( \mathrm{B = \dfrac{\mu_0 I}{2 \pi r}} \)

  • \(\mathrm{\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A}\) is the permeability of free space.
  • The field decreases with distance (\(1/r\)).
  • Field lines are concentric circles around the wire.

Magnetic Field Inside a Current-Carrying Wire:

For a wire of radius \(R\), carrying uniform current distribution:

\( \mathrm{B(r) = \dfrac{\mu_0 I r}{2 \pi R^2}}, \quad (r < R) \)

  • Inside the wire, the field increases linearly with distance from the axis.
  • At the surface (\(r = R\)), this matches the external formula.

Magnetic Field Due to a Circular Current Loop:

At the center of a circular loop of radius \(R\) carrying current \(I\):

\( \mathrm{B = \dfrac{\mu_0 I}{2R}} \)

  • Direction given by the right-hand rule (thumb = current, curl = field).

Properties:

  • The field strength increases with current (\(I\)) and decreases with distance (\(r\)).
  • Magnetic field lines form closed loops around the current.
  • The principle of superposition applies: fields from multiple wires add vectorially.

Example :

A long straight wire carries a current of \(6.0 \, A\). Find the magnetic field at a point \(3.0 \, cm\) from the wire.

▶️ Answer/Explanation

Step 1: Formula: \( \mathrm{B = \dfrac{\mu_0 I}{2 \pi r}} \)

Step 2: Substitute: \( \mathrm{B = \dfrac{(4\pi \times 10^{-7})(6.0)}{2 \pi (0.03)}} \)

Step 3: \( \mathrm{B = 4.0 \times 10^{-5} \, T} \)

Final Answer: The magnetic field is \( \mathrm{4.0 \times 10^{-5} \, T} \), directed circularly around the wire.

Example :

A circular loop of radius \(0.20 \, m\) carries a current of \(3.0 \, A\). Find the magnetic field at the center of the loop.

▶️ Answer/Explanation

Step 1: Formula: \( \mathrm{B = \dfrac{\mu_0 I}{2R}} \)

Step 2: Substitute: \( \mathrm{B = \dfrac{(4\pi \times 10^{-7})(3.0)}{2 (0.20)}} \)

Step 3: \( \mathrm{B = 9.4 \times 10^{-6} \, T} \)

Final Answer: The magnetic field at the center is \( \mathrm{9.4 \times 10^{-6} \, T} \).

Force on Current-Carrying Wires in a Magnetic Field

A wire carrying electric current consists of moving charges (electrons). When placed in a magnetic field, these moving charges experience a magnetic force, which results in a net force on the entire wire.

Mathematical Expression:

The force on a length \(L\) of a straight wire carrying current \(I\) in a magnetic field \(\vec{B}\) is:

\( \mathrm{\vec{F} = I \, \vec{L} \times \vec{B}} \)

  • \(\mathrm{I}\) = current in the wire
  • \(\mathrm{\vec{L}}\) = length vector of the wire in the direction of current
  • \(\mathrm{\vec{B}}\) = magnetic field

Magnitude of the Force:

\( \mathrm{F = I L B \sin \theta} \)

  • \(\theta\) = angle between wire and field direction.
  • Maximum force when wire is perpendicular (\(\theta = 90^\circ\)).
  • No force when wire is parallel to the field (\(\theta = 0^\circ\)).

Direction:

  • Right-hand rule: point thumb along the current, fingers along \(\vec{B}\); palm points in the direction of the force.

Applications:

  • Electric motors: forces on current-carrying wires produce rotational motion.
  • Speakers: wires in magnetic fields vibrate due to alternating current.

Force Between Two Parallel Wires:

  • Two parallel wires carrying currents exert magnetic forces on each other.
  • Force per unit length between two parallel wires separated by distance \(r\):

    \( \mathrm{\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2 \pi r}} \)

  • Currents in the same direction → attraction.
  • Currents in opposite directions → repulsion.

Example :

A \(0.20 \, m\) long wire carries a current of \(5.0 \, A\) perpendicular to a uniform magnetic field of \(0.30 \, T\). Find the force on the wire.

▶️ Answer/Explanation

Step 1: Formula: \( \mathrm{F = I L B \sin \theta} \).

Step 2: Substitute: \( \mathrm{F = (5.0)(0.20)(0.30)(\sin 90^\circ)} \).

Step 3: \( \mathrm{F = 0.30 \, N} \).

Final Answer: The wire experiences a force of \( \mathrm{0.30 \, N} \).

Example :

Two long parallel wires carry currents of \(I_1 = 8.0 \, A\) and \(I_2 = 6.0 \, A\) in the same direction. If the wires are separated by \(0.05 \, m\), find the force per unit length on each wire.

▶️ Answer/Explanation

Step 1: Formula: \( \mathrm{\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2 \pi r}} \).

Step 2: Substitute: \( \mathrm{\dfrac{F}{L} = \dfrac{(4\pi \times 10^{-7})(8.0)(6.0)}{2 \pi (0.05)}} \).

Step 3: \( \mathrm{\dfrac{F}{L} = 1.92 \times 10^{-4} \, N/m} \).

Step 4: Since currents are in the same direction, the force is attractive.

Final Answer: Each wire experiences an attractive force of \( \mathrm{1.92 \times 10^{-4} \, N/m} \).

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